Page 133 - Handbook Of Integral Equations

2.1-3. Kernels Cubic in the Arguments x and t

x
3
20. y(x)+ A x y(t) dt = f(x).
a
Solution:
x
4
4
3

y(x)= f(x) – A x exp 1 A(t – x ) f(t) dt.
4
a
x
2
21. y(x)+ A x ty(t) dt = f(x).
a
Solution:
x
4
4
2

y(x)= f(x) – A x t exp 1 A(t – x ) f(t) dt.
4
a
x
2
22. y(x)+ A xt y(t) dt = f(x).
a
Solution:
x
2
4
4

y(x)= f(x) – A xt exp 1 A(t – x ) f(t) dt.
4
a
x

3
23. y(x)+ A t y(t) dt = f(x).
a
Solution:
x
3
4
4

y(x)= f(x) – A t exp 1 A(t – x ) f(t) dt.
4
a
x

3
24. y(x)+ λ (x – t) y(t) dt = f(x).
a
This is a special case of equation 2.1.34 with n =3.
Solution:
x
y(x)= f(x) – R(x – t)f(t) dt,
a
where

3 1/4
k cosh(kx) sin(kx) – sinh(kx) cos(kx) , k = λ for λ >0,
R(x)= 2
1 s sin(sx) – sinh(sx) , s =(–6λ) 1/4 for λ <0.

2
x

3
3
25. y(x)+ A (x – t )y(t) dt = f(x).
a
This is a special case of equation 2.1.52 with λ =3.
x
3 3
26. y(x) – A 4x – t y(t) dt = f(x).
a
This is a special case of equation 2.1.55 with λ = 1 and µ =3.
x
3
2
27. y(x)+ A (xt – t )y(t) dt = f(x).
a
This is a special case of equation 2.1.49 with λ =2.
© 1998 by CRC Press LLC
© 1998 by CRC Press LLC
Page 112